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Khách

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\(\Leftrightarrow-x^3-x⋮x^2-2\)

\(\Leftrightarrow-x^3+2x-3x⋮x^2-2\)

\(\Leftrightarrow-3x^2⋮x^2-2\)

\(\Leftrightarrow x^2-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{1;-1;2;-2\right\}\)

4 tháng 2 2019

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+3+...+x\right)\)

\(B=1+\frac{1}{2}\left(1+2\right)\cdot2:2+\frac{1}{3}\left(1+3\right)\cdot3:2+...+\frac{1}{x}\left(1+x\right)\cdot x:2\)

\(B=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+x}{2}\)

\(B=1+\frac{\left(1+1+...+1\right)+\left(2+3+...+x\right)}{2}\)

De B = 115

=> \(\frac{\left(1+1+...+1\right)+\left(2+3+...+x\right)}{2}=114\)

=> (1 + 1 + ... + 1) + (2 + 3 + ... + x) = 228

den day chju :v

4 tháng 2 2019

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.............+\frac{1}{x}\left(1+2+3+............+x\right)\)

\(=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}\frac{3.4}{2}+...........+\frac{1}{x}\frac{x\left(x+1\right)}{2}\)

\(=\frac{1}{2}\left(2+3+4+.............+\left(x+1\right)\right)\)

\(=\frac{1}{2}\frac{\left[\left(x+1\right)+2\right]x}{2}\)

\(=\frac{1}{4}\left(x+3\right)x\)

\(B=115\Leftrightarrow\frac{1}{4}.x\left(x+3\right)=115\)

\(\Leftrightarrow x\left(x+3\right)=115.4\)

\(\Leftrightarrow x\left(x+3\right)=20.23\)

\(\Leftrightarrow x=20\)